Question ID: #589
If the function $f(x)=2x^{3}-9ax^{2}+12a^{2}x+1$, where $a>0$, attains its local maximum and local minimum values at $p$ and $q$, respectively, such that $p^{2}=q$, then $f(3)$ is equal to:
- (1) 55
- (2) 10
- (3) 23
- (4) 37
Solution:
Find the derivative of $f(x)$:
$$f'(x) = 6x^2 – 18ax + 12a^2$$
Set $f'(x) = 0$ to find critical points:
$$6(x^2 – 3ax + 2a^2) = 0$$
$$(x – a)(x – 2a) = 0$$
The critical points are $x = a$ and $x = 2a$.
Since $a > 0$, we check the second derivative $f”(x) = 12x – 18a$:
$$f”(a) = 12a – 18a = -6a < 0 \quad (\text{Local Max at } p = a)$$ $$f''(2a) = 24a - 18a = 6a > 0 \quad (\text{Local Min at } q = 2a)$$
Given the condition $p^2 = q$:
$$a^2 = 2a$$
Since $a > 0$, we divide by $a$:
$$a = 2$$
Now substitute $a=2$ back into the original function:
$$f(x) = 2x^3 – 18x^2 + 48x + 1$$
Calculate $f(3)$:
$$f(3) = 2(27) – 18(9) + 48(3) + 1$$
$$f(3) = 54 – 162 + 144 + 1$$
$$f(3) = 37$$
Ans. (4)
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